poj 2442【堆應用+STL Heap】

本題要維護兩個大根堆，用來迴圈維護。。。注意STL heap裡的複雜度

#include <vector>#include <list>#include <map>#include <set>#include <queue>#include <string.h>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <limits.h>using namespace std;int lowbit(int t){return t&(-t);}int countbit(int t){return (t==0)?0:(1+countbit(t&(t-1)));}int gcd(int a,int b){return (b==0)?a:gcd(b,a%b);}#define LL long long#define PI acos(-1.0)#define N  2010#define MAX INT_MAX#define MIN INT_MIN#define eps 1e-8#define FRE freopen("a.txt","r",stdin)int a[N],b[N];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        int i,j,k;        for(i=0;i<m;i++)        scanf("%d",&a[i]);        make_heap(a,a+m);//複雜度O(n)        int num;        for(i=1;i<n;i++){            for(j=0;j<m;j++){                scanf("%d",&num);                if(i&1){                    if(j==0){                        for(k=0;k<m;k++){                            b[k]=a[k]+num;                        }                    }                    else{                        for(k=0;k<m;k++){                            if(num+a[k]<b[0]){                                pop_heap(b,b+m);//複雜度O(log(n))                                b[m-1]=num+a[k];                                push_heap(b,b+m);//把b[m-1]推進去堆裡再做成一個堆,複雜度O(log(n))                            }                        }                    }                }                else{                    if(j==0){                        for(k=0;k<m;k++){                            a[k]=b[k]+num;                        }                    }                    else{                        for(k=0;k<m;k++){                            if(num+b[k]<a[0]){                                pop_heap(a,a+m);                                a[m-1]=num+b[k];                                push_heap(a,a+m);                            }                        }                    }                }            }        }        if(!(n&1)){            sort(b,b+m);            for(i=0;i<m;i++)            printf("%d ",b[i]);            puts("");        }        else{            sort(a,a+m);            for(i=0;i<m;i++)            printf("%d ",a[i]);            puts("");        }    }    return 0;}